Problem

A gray–iron casting product is been molded with a green sand stacking molding process; which is known to produce pin–hole defects. The current yield is 76%. The plant manager wants to increase the plant’s yield.

Pouring molten metal into the ladle

The pin–holes are associated with the molten metal temperature at the furnace, at the time of the pouring process. The acceptable minimum is 1127 C. However, the gray iron alloy is known to be stable at 1204 C.

Pouring molten metal into the stack of molds

An experiment was conducted to verify

\[yield=f(temp)+\epsilon\] at three temperatures. The runs’ results are shown in table 1.

1127 1178 1204
73.4 74.4 78.9
76.3 74.2 79.3
76.9 72.7 77.9
77.9 73.5 78.9
77.7 71.4 79.5

\(Yield=f(T)\)?

Data

data.set <- data.frame(yield=
                     c(73.4,76.3,76.9,77.9,77.7,
                       74.4,74.2,72.7,73.5,71.4,
                       78.9,79.3,77.9,78.9,79.5),
                     temp=as.factor(
                       c(rep('1127',5),
                         rep('1178',5),
                         rep('1204',5)
                         )))
                     
data.set
##    yield temp
## 1   73.4 1127
## 2   76.3 1127
## 3   76.9 1127
## 4   77.9 1127
## 5   77.7 1127
## 6   74.4 1178
## 7   74.2 1178
## 8   72.7 1178
## 9   73.5 1178
## 10  71.4 1178
## 11  78.9 1204
## 12  79.3 1204
## 13  77.9 1204
## 14  78.9 1204
## 15  79.5 1204

Descriptive statistics

#Boxplot
fn <- yield~temp
boxplot (fn,data.set)

library(gplots)
plotmeans(fn,data.set,ci.label=T,n.label=F)

Variance Homogeneity

bartlett.test(fn,data.set)
## 
##  Bartlett test of homogeneity of variances
## 
## data:  yield by temp
## Bartlett's K-squared = 3.6215, df = 2, p-value = 0.1635
fligner.test(fn,data.set)
## 
##  Fligner-Killeen test of homogeneity of variances
## 
## data:  yield by temp
## Fligner-Killeen:med chi-squared = 1.9389, df = 2, p-value = 0.3793

Statistical Model—ANOVA 1w.

The potential relation of the yield and the temperature are explored as a fixed effect model:

\[Y_{ij}=\mu+\tau_i+\epsilon_{ij}\]

Where:

  • \(Y_{ij}\)—The runs’ results, \(i\in(1,...,k)\) treatments (levels) with \(j\)-observations;
  • \(\mu\)—The mean of means;
  • \(\tau\)—The fixed treatment’s effects;
  • \(\epsilon_{ij}\)—The runs’ errors.
fit_1 <- aov(fn,data.set)
summary(fit_1)
##             Df Sum Sq Mean Sq F value   Pr(>F)    
## temp         2  80.55   40.27   23.32 7.34e-05 ***
## Residuals   12  20.72    1.73                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
TukeyHSD(fit_1)
##   Tukey multiple comparisons of means
##     95% family-wise confidence level
## 
## Fit: aov(formula = fn, data = data.set)
## 
## $temp
##            diff        lwr        upr     p adj
## 1178-1127 -3.20 -5.4173783 -0.9826217 0.0060433
## 1204-1127  2.46  0.2426217  4.6773783 0.0297891
## 1204-1178  5.66  3.4426217  7.8773783 0.0000517

Power analysis

power.anova.test(groups = 3, n = 5, between.var = 40.27, within.var = 1.73, power = NULL)
## 
##      Balanced one-way analysis of variance power calculation 
## 
##          groups = 3
##               n = 5
##     between.var = 40.27
##      within.var = 1.73
##       sig.level = 0.05
##           power = 1
## 
## NOTE: n is number in each group

Marginal effects

require(margins)
cplot(fit_1)
##   xvals yvals    upper    lower
## 1  1127 76.44 77.59189 75.28811
## 2  1178 73.24 74.39189 72.08811
## 3  1204 78.90 80.05189 77.74811

Graphical analysis \((\tau_i)\)

## $grandsum
##     Grandmean        df.bet       df.with        MS.bet       MS.with 
##         76.19          2.00         12.00         40.27          1.73 
##        F.stat        F.prob SS.bet/SS.tot 
##         23.32          0.00          0.80 
## 
## $stats
##       Size Contrast Coef Wt'd Mean  Mean Trim'd Mean Var. St. Dev.
## X1178    5         -2.95     73.24 73.24       73.47 1.50     1.23
## X1127    5          0.25     76.44 76.44       76.97 3.30     1.82
## X1204    5          2.71     78.90 78.90       79.03 0.38     0.62